2022 CHIJ Phy Prelims P1&2 Ans

CHIJ SECONDARY (TOA PAYOH) PRELIMINARY EXAMINATION 2022 SECONDARY 4(EXPRESS) PHYSICS 6091/01 Paper 1 1 D 11 B 21 B 31 B 2 A 12 B 22 D 32 D 3 B 13 C 23 A 33 C 4 C 14 B 24 A 34 C 5 B 15 D 25 A 35 D 6 B 16 D 26 C 36 C 7 C 17 C 27 D 37 A 8 B 18 D 28 B 38 C 9 B 19 A 29 C 39 A 10 C 20 A 30 B 40 D Paper 2 6091/02 Qn Answers Mark 1(a) The distance travelled per unit time is increasing. 1 1(b) average speed = total distance / total time = 80 cm / 2 s = 40 cm/s or 0.4 m/s 1 1 1(c) Air resistance is increasing. Weight is constant/ remain unchanged. 1 1 2(a) - Gradient on a distance-time graph represents speed - Gradient of P is greater than gradient of Q 1 1 2(b) 120 km 1 2(c) Speed = distance / time = 120 km / 1.5 hours = 80 km/h 1 1 2(d) 1 3(a) - Appropriate scale - Correct resultant force indicated with double arrow - Correct addition of forces drawn - Range: 135kN to 165 kN 1 1 1

3(b) - The resultant force of the two tugboats is equal and opposite to the friction between ship and water. - Resultant force on ship is 0 N. 1 1 4(a) Volume = 3.6x0.35x0.025 = 0.0315 m3 Density = mass / volume = 23 / 0.0315 = 730 kg/m3 1 1 4(b)(i) W = mg = 23 x 10 = 230 N 1 1 4(b)(ii) Moment = force x perpendicular distance = 230 x 1.3 = 299 Nm 1 1 4(c) - When painter moves further to the right, his weight will create a clockwise moment about the right-hand support. - This clockwise moment is greater than the anti-clockwise moment due to the weight of the plank about the same pivot. 1 1 5(a) - Chemical potential energy Kinetic energy + Gravitational potential energy - Chemical potential energy is decreasing, kinetic energy is constant, gravitational potential energy is increasing. OR Chemical potential energy Gravitational potential energy + thermal energy + sound 1 1 5(b) All the chemical potential energy input by the cyclist is converted to thermal energy when the cyclist stops. 1 5(c) GPE = mgh 5700 = 60 x 10 x h h = 9.5 m 1 1 6(a) less same less greater same 1 1 1 1 1 7(a) Three errors: 1. Ultra-sound should be ultraviolet instead 2. The positions of microwaves and infra-red waves should be interchange. 3. Visible light is missing. It should be between ultraviolet and infra-red waves. 1 1 1 7(b) X-rays is used to check for internal cracks in metal structures. 1 8(a) - Bring positively charged rod near but not touching sphere. - Connect the earth connection to the sphere. - Remove the earth connection. - Remove the positively charged rod. 1 1 1 8(b) - When positively charged rod is brought near the sphere, electrons on sphere will be attracted to the left side of sphere. - When earth connection is connected to sphere, electrons will be attracted from the earth into the sphere. 1 1 8(c) Q = It I = Q/t = 2.5x10-9 / 4.0x10-3 = 625 nA 1 1 9(a) - Soft-iron is magetised into a strong electromagnet - Iron lever is attracted by the strong electromagnet and starts to turn clockwise about the pivot - Spring will contract and pull the springy metal towards the reset button - Contact will open 1 1 1 1 9(b) Increase the number of turns per unit length in the coil 1 END OF SECTION A

10(a) - As the turbine rotates, the coil will experience a rate of change of magnetic flux around it. - By Faraday’s law of electromagnetic Induction, an induced e.m.f. is formed on the coil. 1 2 10(a)(ii) - Amplitude increased to 0.1 V - Period reduced by ½ - time interval between waveform is reduced from 60 ms to 30 ms. 1 1 1 10(c) Turns ratio = Ns / Np = Vs / Vp = 400k / 25k = 16 1 10(d) Vs / Vp = Ip / Is 16 = 12 x 103 / Is Is = 750 A 1 1 Power input = power output Ip Vp = Is Vs 10(e) Reduce energy loss through thermal on the transmission cables. 1 Due to lower current transmitted. 11(a) Input transducers are electronic devices that convert non- electrical energy to electrical energy while output transducers convert electrical energy to non-electrical energy. 1 (b)(i) Resistance at 40 °C = 2.5 kΩ 1 (b)(ii) Graph is not a straight line / Resistance of the sensor changes non-linearly 1 (c) I = V / R = 20.0 / [10 x 103] = 2 mA 1 1 (d)(i) V = I R = 2 x 10-3 x 7500 = 15 V 1 (d)(ii) Voutput = R/[R+RT] x 20.0 V 6.0 = 7500/[7500+RT] x 20.0 V RT = 17.5 k From graph, minimum temperature = 0 ˚C 10.0 = 7500/[7500+RT] x 20.0 V RT = 7.5 k From the graph, maximum temperature = 15 ˚C 1 1 1 (d)(iii) To switch on a heater when the temperature drops below 15 °C. 1 12 Either 12(a) Effective resistance between AB = [1/(1/10+1/10) + 1/(1/10+1/10+1/10+1/10)] = 7.5 Ω 1 1 12(b)(i) - E.m.f is defined as the amount of work done by an electrical source in driving one unit charge around a complete circuit. - Potential difference is defined as the amount of work done in driving one unit charge across a component in the electrical circuit. 1 1

12(b)(ii) Potential difference across AB = 24 – 6 = 18 V 1 12(c)(i) V = IR I = V / R = 18 / 7.5 = 2.4 A 1 1 12(c)(ii) R = V / I = 6.0 / 2.4 = 2.5 Ω 1 12(d) - In the event of an electrical fault, the earth wire will conduct the large fault current safely to the ground. - This will prevent user from getting an electric shock if user accidentally touches the metal casing 1 1 12OR (a)(i) - Correct pair of rays drawn - Correct F labelled 1 1 (a)(ii) - Correct pair of rays drawn - Correct F labelled 1 1 (b) Image is upright. 1 (c) Overhead Projector It is able to project a big image of a small object onto a screen. 1 1 (d) Object should be placed exactly 2 focal length away from lens. 1 (e) 1. The incident ray, the normal and the refracted ray at the point of incidence all lie on the same plane. 2. For 2 given media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. 1 1 END OF PAPER 2